Question: Let $y=(14+5x-3x^2)^{^{{\scriptsize\dfrac34}}}$. Evaluate $\dfrac{dy}{dx}$ at $x=1$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac32$ (Choice B) B $-\dfrac38$ (Choice C) C $2$ (Choice D) D $\dfrac34$
Let's start by finding the expression for $\dfrac{dy}{dx}$. Then, we can evaluate it at $x=1$. $(14+5x-3x^2)^{^{{\scriptsize\dfrac34}}}$ is a power function with a rational exponent, but its argument isn't simply $x$. Therefore, it defines a composite power function. In other words, suppose $u(x)=14+5x-3x^2$, then $y=[u(x)]^{^{{\scriptsize\dfrac34}}}$. $\dfrac{dy}{dx}$ can be found using the following identity: $\dfrac{d}{dx}\left[[u(x)]^{^{{\scriptsize\dfrac34}}}\right]=\dfrac{3}{4}[u(x)]^{^{-\scriptsize\dfrac{1}{4}}}u'(x)$ [Why is this identity true?] Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}(14+5x-3x^2)^{^{{\scriptsize\dfrac34}}} \\\\ &=\dfrac{d}{dx}[u(x)]^{^{{\scriptsize\dfrac34}}}&&\gray{\text{Let }u(x)=14+5x-3x^2} \\\\ &=\dfrac{3}{4}[u(x)]^{^{-\scriptsize\dfrac{1}{4}}}u'(x) \\\\ &=\dfrac{3}{4}[14+5x-3x^2]^{^{-\scriptsize\dfrac{1}{4}}}(5-6x)&&\gray{\text{Substitute }u(x)\text{ back}} \end{aligned}$ Now let's evaluate $\dfrac{dy}{dx}$ at $x= 1$. $\begin{aligned} &\phantom{=}\dfrac{3}{4}\Bigl(14+5( 1)-3( 1)^2\Bigr)^{^{-\scriptsize\dfrac{1}{4}}}\cdot\Bigl(5-6( 1)\Bigr) \\\\ &=\dfrac{3}{4}\cdot 16^{^{-\scriptsize\dfrac{1}{4}}}\cdot(-1) \\\\ &=\dfrac{3}{4}\cdot \dfrac12\cdot (-1) \\\\ &=-\dfrac{3}{8} \end{aligned}$ In conclusion, the value of $\dfrac{dy}{dx}$ at $x=1$ is $-\dfrac{3}{8}$.